- A student was asked to prove, for \(p \in \mathbb { N }\), that
"if \(p ^ { 3 }\) is a multiple of 3 , then \(p\) must be a multiple of 3 "
The start of the student's proof by contradiction is shown in the box below.
Assumption:
There exists a number \(p , p \in \mathbb { N }\), such that \(p ^ { 3 }\) is a multiple of 3 , and \(p\) is NOT a multiple of 3
Let \(p = 3 k + 1 , k \in \mathbb { N }\).
$$\text { Consider } \begin{aligned}
p ^ { 3 } = ( 3 k + 1 ) ^ { 3 } & = 27 k ^ { 3 } + 27 k ^ { 2 } + 9 k + 1
& = 3 \left( 9 k ^ { 3 } + 9 k ^ { 2 } + 3 k \right) + 1 \quad \text { which is not a multiple of } 3
\end{aligned}$$
- Show the calculations and statements that are required to complete the proof.
- Hence prove, by contradiction, that \(\sqrt [ 3 ] { 3 }\) is an irrational number.